Two force always involved in using a machine are

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

Calculate the change in entropy as 0.3071 kg of ice at 273.15 K melts. (The latent heat of fusion of water is 333000 J / kg)

Ostrovityanka [42]

Answer:

374.39 J/K

Explanation:

Entropy: This can be defined as the degree of disorder or randomness of a substance.

The S.I unit of entropy is J/K

ΔS = ΔH/T ..................................... Equation 1

Where ΔS = entropy change, ΔH = Heat change, T = temperature.

ΔH = cm................................... Equation 2

Where,

c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.

Substitute into equation 2

ΔH = 333000×0.3071

ΔH = 102264.3 J.

Also, T = 273.15 K

Substitute into equation 1

ΔS = 102264.3/273.15

ΔS = 374.39 J/K

Thus, The change in entropy = 374.39 J/K

3 0

3 years ago

Which unit is used to measure force

Ymorist [56]

Newtons (N) measure force

3 0

3 years ago

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4. What is the magnitude and direction of the gravitational force that acts on a man

Cloud [144]

The gravitational force acting on the man is 800 N towards the Earth's centre

Explanation:

The weight of an object on the Earth is exactly the gravitational force exerted by the Earth on the object.

The gravitational force exerted by the Earth on an object located at the Earth's surface is given by:

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

R is the radius of the Earth

And the direction of the force is towards the Earth's centre.

Since G, M and R are constant, they are grouped into a single constant called g, acceleration of gravity:

$g=\frac{GM}{R^2}$

therefore the gravitational force can be rewritten as

$F=mg$

And this is the usual equation that we use to calculate the weight of an object.

Therefore, weight and gravitational force acting on an object on Earth are the same thing: so, the gravitational force acting on the man is equal to his weight, 800 N, and it acts towards the Earth's center.

Learn more about weight and gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

Consider this situation: A rope is used to lift an actor upward off the stage. Of the forces listed, identify which act upon the

Katarina [22]

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6 0

4 years ago

Read 2 more answers